what is the distance between P and the position of the ball strike the wall?u is 20ms-1 and the angle is 40 degree?

(i)
Consider the horizontal motion (uniform velocity motion):
s(x) = v(x) t
Time taken, t = s(x)/v(x) = 60/(20 cos40°) s = 3.916 s ≈ 3.92 s

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(ii)
Consider the vertical motion (uniform acceleration motion):
Take g = 9.8 m s⁻²
Take all upward quantities to be positive, and downward quantities to be negative.

Initial velocity, u(y) = 20 sin40° m s⁻¹
Acceleration, a(y) = -9.8 m s⁻²
Time taken, t = 3.916 s

s(y) = u(y) t + (1/2) a t²
s(y) = (20 sin40°) (3.916) + (1/2) (-9.8) (3.916) m
s(y) = -24.8 m

Distance between P and the position the ball strikes the wall = 24.8 m
(The position is below P.)

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(iii)
Consider the vertical motion:
v(y) = u(y) + at
v(y) = 20 sin40° + (-9.8) (3.916)
Vertical velocity when the ball strikes the wall, v(y) = -25.52 m s⁻¹
(The vertical velocity is pointing downward.)

Consider the horizontal motion:
v(x) = 20 cos40° m s⁻¹

The speed of the ball when it strikes the wall
= √[v(x)² + v(y)²]
= √[(20 cos40°)² + (-25.52)²] m s⁻¹
= 29.8 m s⁻¹

horizontal component of velocity, v𝑥 = 20 cos40 = 15.3 m/s
(i) Time to hit the wall, t = (60m)/(15.32 m/s) = 3.92 s

(ii) vertical distance s = ut+½gt²
vertical component of velocity, v𝑦 = 20 sin40 = 12.9 m/s
s = (12.9)(3.92)+½(-9.81)(3.92)²
s = -24.8 

The distance between P and the location where ball strikes is 24.8 m (below the initial level) 

(iii) horizontal component of velocity after 3.92s 

v = u+at
v = 12.9 + (-9.81)(3.92)
v = -25.6 m/s

speed of ball when it strikes the wall = √(-25.6)² + (15.32)²) = 29.8 m/s